HDU2192 MagicBuilding【水题】

MagicBuilding

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2987 Accepted Submission(s): 1346

Problem Description
As the increase of population, the living space for people is becoming smaller and smaller. In MagicStar the problem is much worse. Dr. Mathematica is trying to save land by clustering buildings and then we call the set of buildings MagicBuilding. Now we can treat the buildings as a square of size d, and the height doesn't matter. Buildings of d1,d2,d3....dn can be clustered into one MagicBuilding if they satisfy di != dj(i != j).
Given a series of buildings size , you need to calculate the minimal numbers of MagicBuildings that can be made. Note that one building can also be considered as a MagicBuilding.
Suppose there are five buildings : 1, 2, 2, 3, 3. We make three MagicBuildings (1,3), (2,3), (2) .And we can also make two MagicBuilding :(1,2,3), (2,3). There is at least two MagicBuildings obviously.

Input
The first line of the input is a single number t, indicating the number of test cases.
Each test case starts by n (1≤n≤10^4) in a line indicating the number of buildings. Next n positive numbers (less than 2^31) will be the size of the buildings.

Output
For each test case , output a number perline, meaning the minimal number of the MagicBuilding that can be made.

Sample Input
2
1
2
5
1 2 2 3 3

Sample Output
1
2

问题链接：HDU2192 MagicBuilding
问题简述：（略）
问题分析：
    统计序列中出现次数最多数的次数，大水题。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* HDU2192 MagicBuilding */

#include <iostream>
#include <algorithm>
#include <stdio.h>

using namespace std;

const int N = 10000;
int a[N];

int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);

        sort(a, a + n);

        int cnt = 1, maxcnt = 1, cur = a[0];
        for(int i = 1; i < n; i++) {
            if(a[i] == cur) cnt++;
            else cur = a[i], cnt = 1;
            maxcnt = max(maxcnt, cnt);
        }

        printf("%d\n", maxcnt);
    }

    return 0;
}